Problem: Rewrite the function by completing the square. $f(x)=x^{2}-6x+43$ $f(x)=(x+$
We want to complete $x^2{-6}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-6}}{2}\right)^2={9}$ to it: $x^2{-6}x+{9}=(x-3)^2$ In order to keep the expression equivalent, we add and subtract ${9}$, not forgetting the expression's constant term, $43$ : $\begin{aligned} f(x)&=x^2-6x+43 \\\\ &=x^2-6x+{9}+43-{9} \\\\ &=(x-3)^2+43-9 \\\\ &=(x-3)^2+34 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 3)^2 + 34$ This is equivalent to $f(x)=(x+{-3})^2+34$